Các bài toán đại số trong các kì thi Olympic

(a
2
+ b
2
+ c
2
+ d
2
)(x
2
+ y
2
+ z
2
+ t
2
) = (ax − by − cz − dt)
2
+ (bx +
ay −dz + ct)
2
+ (cx + dy + az −dt)
2
+ (dx − cy + bz + at)
2
ax−by−cz−dt = 0 bx+ay−dz+ct = 0
cx + dy + az − dt = 0 dx − cy + bz + at = 0
a = b = c = d = 0 x = y = z = t = 0
(a
2

+b
2
+c
2
)(x
2
+y
2
+z
2
)−(ax+by+cz)
2
= (bx−ay)
2
+(cy−bz)
2
+(az−cx)
2
(a
2
1
+ a
2
2
+ ··· + a
2
n
)(b
2
1

+ b
2
2
+ ··· + b
2
n
) − (a
1
b
1
+ a
2
b
2
+ ··· + a
n
b
n
)
2
=
= (a
1
b
2
− a
2
b
1
)

2
+ (a
1
b
3
− a
3
b
1
)
2
+ ··· + (a
n−1
b
n
− a
n
b
n−1
)
2
n(a
2
1
+ a
2
2
+ ··· + a
2
n

) = (a
1
+ a
2
+ ··· + a
n
)
2
a
1
= a
2
= ··· = a
n
(x − y)
2
+ (y − z)
2
+ (z − x)
2
= (y + z −
2x)
2
+ (z + x − 2y)
2
+ (x + y −2z)
2
x = y = z
(a
2

− b
2
)
2
+ (2ab)
2
= (a + b)
2
(6a
2
−4ab+4b
2
)
3
= (3a
2
+5ab−5b
2
)
3
+(4a
2
−4ab+6b
2
)
3
+(5a
2
−5ab−3b
2

)
3
(p
2
− q
2
)
4
+ (2pq + q
2
)
4
+ (2pq + p
2
)
4
= 2(p
2
+ pq + q
2
)
4
X
2
+ XY + Y
2
= Z
3
X = q
3

+ 3pq
2
− p
3
Y =
−3pq(p + q) Z = p
2
+ pq + q
2
(3a + 3b)
k
+ (2a + 4b)
k
+ a
k
+ b
k
= (3a + 4b)
k
+ (a + 3b)
k
+ (2a + b)
k
k = 1, 2, 3
x + y + z = 0
(ix −ky)
n
+ (iy −kz)
n
+ (iz −kx)

n
= (iy −kx)
n
+ (iz −ky)
n
+ (ix −kz)
n
n = 0, 1, 2, 4 i i
2
= −1
x
n
+ (x + 3)
n
+ (x + 5)
n
+ (x + 6)
n
+ (x + 9)
n
+ (x +
10)
n
+ (x + 12)
n
+ (x + 15)
n
= (x + 1)
n
+ (x + 2)

n
+ (x + 4)
n
+ (x + 7)
n
+
(x + 8)
n
+ (x + 11)
n
+ (x + 13)
n
+ (x + 14)
n
n = 0, 1, 2, 3
(a +b + c + d)
2
+ (a + b −c −d)
2
+ (a + c −b −d)
2
+ (a + d −b −c)
2
=
4(a
2
+ b
2
+ c
2

+ d
2
)
(a
2
− b
2
+ c
2
− d
2
)
2
+ 2(ab − bc + dc + ad)
2
= (a
2
+ b
2
+ c
2
+ d
2
)
2
−
2(ab − ad + bc + dc)
2
(a
2

−c
2
+2bd)
2
+(d
2
−b
2
+2ac)
2
= (a
2
−b
2
+c
2
−d
2
)
2
+2(ab−bc+dc+ad)
2
(a+b+c)
4
+(a+b−c)
4
+(a−b+c)
4
+(−a+b+c)
4

= 4(a
4
+b
4
+c
4
)+24(a
2
b
2
+b
2
c
2
+c
2
a
2
)
s = a + b + c = 2p

sym
s(s − 2b)(s − 2c) = (s − 2a)(s − 2b)(s − 2c) + 8abc

sym
a(p − a)
2
= abc − 2(p − a)(p − b)(p − c )
s = a + b + c 2δ = a
2

+ b
2
+ c
2

sym
(δ
2
− a
2
)(δ
2
− b
2
) = 4s(s − a)(s − b)(s − c)
a + b + c = 0 a
3
+ b
3
+ c
3
= 3abc
a
3
+ b
3
+ c
3
− 3abc = (a + b + c)(a
2

+ b
2
+ c
2
− ab − bc − ca)
a, b, c
(a + b + c)
3
−

sym
(a + b − c)
3
(a − b)
3
+ (b − c)
3
+ (c − a)
3
= 3(a − b)(b − c)(c − a)
[(a − b)
2
+ (b − c)
2
+ (c − a)
2
]
2
= 2[(a − b)
4

+ (b − c)
4
+ (c − a)
4
]
a + b + c = 0
• 2(a
4
+ b
4
+ c
4
) = (a
2
+ b
2
+ c
2
)
2
•
a
5
+b
5
+c
5
5
= abc ·
a

2
+b
2
+c
2
2
•
a
3
+b
3
+c
3
3
·
a
2
+b
2
+c
2
2
=
a
5
+b
5
+c
5
5

•
a
7
+b
7
+c
7
7
=
a
2
+b
2
+c
2
2
·
a
5
+b
5
+c
5
5
•
a
7
+b
7
+c

7
7
=
a
3
+b
3
+c
3
3
·
a
4
+b
4
+c
4
4
·
2n a
1
, a
2
, . . . , a
n
b
1
, b
2
, . . . , b

n
s
k
= a
1
b
1
+ a
2
b
2
+
··· + a
k
b
k
k = 1, 2, , n
n

k=1
a
k
b
k
=
n

k=1
(a
k

− a
k+1
)s
k
n
a
1
+ a
2
+ ··· + a
n
=
n
2
s
n

k=1
(s − a
k
)
2
=
n

k=1
a
2
k
Ax

2
+ 2Bxy + Cy
2
x = αu + β v y = γu + δv
Mu
2
+ 2Nuv + P v
2
N
2
−MP = (B
2
−AC)(αδ −βγ)
2
2n a
1
, a
2
, . . . , a
n
b
1
, b
2
, . . . , b
n
a
i
+ b
i

= 1
a =
a
1
+ a
2
+ ··· + a
n
n
b =
b
1
+ b
2
+ ··· + b
n
n
n

k=1
a
k
b
k
= nab − (a
1
− a)
2
− (a
2

− a)
2
− ··· − (a
n
− a)
2
1 −
1
2
+
1
3
−
1
4
+ ··· +
1
2n − 1
−
1
2n
=
1
n + 1
+
1
n + 2
+ ··· +
1
2n

(1+
1
x − 1
)(1−
1
2x − 1
)(1+
1
3x − 1
) ···(1+
1
(2n − 1)x − 1
)(1−
1
2nx − 1
) =
=
(n + 1)x
(n + 1)x − 1
·
(n + 2)x
(n + 2)x − 1
···
(n + n)x
(n + n)x − 1
x
3
= (x ·
x
3

− 2y
3
x
3
+ y
3
)
3
+ (y ·
2x
3
− y
3
x
3
+ y
3
)
3
2
x
2
− 1
+
4
x
2
− 4
+
6

x
2
− 9
+ ··· +
20
x
2
− 100
= 11

1
(x − 1)(x + 10)
+
1
(x − 2)(x + 9)
+ ··· +
1
(x − 10)(x + 1)

a
b
=
c
d
ab
cd
=
(a + b)
2
(c + d)

2
x =
a − b
a + b
; y =
b − c
b + c
; z =
c − a
c + a
(1 + x)(1 + y)(1 + z) = (1 − x)(1 − y)(1 − z)
(a + b + c + d)(a − b − c + d) = (a − b + c − d)(a + b − c − d)
a
c
=
b
d
ax + by + cz = 0
ax
2
+ by
2
+ cz
2
bc(y −z)
2
+ ca(z −x)
2
+ ab(x − y)
2

=
1
a + b + c
x
2
y
2
z
2
a
2
b
2
+
(x
2
− a
2
)(y
2
− a
2
)(z
2
− a
2
)
a
2
(a

2
− b
2
)
+
(x
2
− b
2
)(y
2
− b
2
)(z
2
− b
2
)
b
2
(b
2
− a
2
)
= x
2
+ y
2
+ z

2
− a
2
− b
2
S
k
=
a
k
(a − b)(a − c)
+
b
k
(b − c)(b − a)
+
c
k
(c − a)(c − b)
S
−2
=
1
abc
· (
1
a
+
1
b

+
1
c
); S
−1
=
1
abc
; S
0
= S
1
= 0; S
2
=
a + b + c; S
4
= ab + bc + ca + a
2
+ b
2
+ c
2
; S
5
= a
3
+ b
3
+ c

3
+ a
2
b + ab
2
+
b
2
c + bc
2
+ c
2
a + ca
2
S
k
=

cyclic
a
k
(a − b)(a − c)(a − d)
S
0
= S
1
= S
2
= 0; S
3

= 1; S
4
= a + b + c + d
S
k
=

cyclic
a
k
(a + b)(a + c)
(a − b)(a − c)
S
0
, S
1
, S
2
, S
3
, S
4

cyclic
ab
(c − x)(c − y)(c − z)
(c − a)(c − b)
= abc − xyz

cyclic

a
2
b
2
c
2
(a − d)(b − d)(c − d)
= abc + bcd + cda + dab

cyclic
a
k
(a − b)(a − c)(x − a)
k = 1, 2

cyclic
b + c + d
(a − b)(a − c)(a − d)(a − x)
=
x − a − b − c − d
(x − a)(x − b)(x − c)(x − d)

cyclic
a
k
(x − b)(x − c)
(a − b)(a − c)
= x
k
k = 0, 1, 2

a + b + c = 0
(
a − b
c
+
b − c
a
+
c − a
b
)(
c
a − b
+
a
b − c
+
b
c − a
) = 9
a − b
a + b
+
b − c
b + c
+
c − a
c + a
+
a − b

a + b
·
b − c
b + c
·
c − a
c + a
= 0

cyclic
b − c
(a − b)(a − c)
= 2

sym
1
a − b

sym
b
2
+ c
2
− a
2
2bc
= 1
1
−1
1

a
+
1
b
+
1
c
=
1
a + b + c
n
1
a
n
+
1
b
n
+
1
c
n
=
1
a
n
+ b
n
+ c
n

bz + cy
x(−ax + by + cz)
=
cx + az
y(ax − by + cz)
=
ay + bx
z(ax + by −cz)
x
a(b
2
+ c
2
− a
2
)
=
y
b(c
2
+ a
2
− b
2
)
=
z
c(a
2
+ b

2
− c
2
)
a + b + c = x + y + z =
x
a
+
y
b
+
z
c
= 0
xa
2
+ by
2
+ cz
2
= 0
a
3
+b
3
+c
3
= (b+c)(c +a)(a+b) (b
2
+c

2
−a
2
)x = (c
2
+a
2
−b
2
)y =
(a
2
+ b
2
− c
2
)z
x
3
+ y
3
+ z
3
= (x + y)(y + z)(z + x)
1
x
+
1
y
=

1
z
(z −x)
2
+ z
2
(z −y)
2
+ z
2
=
x
2
y
2
b − c
1 + bc
,
c − a
1 + ca
,
a − b
1 + ab

cyclic
a
k
(x − b)(x − c)(x − d)
(a − b)(a − c)(a − d)
= x

k
k = 0, 1, 2, 3
x =
3
√
4 +
3
√
2 + 1
(1 +
1
x
)
3
• (a + b + c)(bc + ca + ab) = abc + (b + c)(c + a)(a + b)
• (a
2
− 1)(b
2
− 1)(c
2
− 1) + (a + bc)(b + ca)(c + ab) = (abc + 1)(a
2
+
b
2
+ c
2
+ 2abc − 1)
• (b + c − a)

3
+ (c + a −b)
3
+ (a + b −c)
3
− 4(a
3
+ b
3
+ c
3
− 3abc) =
3(b + c − a)(c + a − b)(a + b − c)
•

cyclic
a
4
(b
2
− c
2
) = (

cyclic
a
2
(b − c))(a + b)(b + c)(c + a)
• a
5

+ b
5
− (a + b)
5
= −5ab(a
2
+ ab + b
2
)
• (a + b)
7
− a
7
− b
7
= 7ab(a + b)(a
2
+ ab + b
2
)
2
xy + yz + zx = 0

sym
(x + y)
2
+ 24x
2
y
2

z
2
=

sym
x
4
(y + z)
2
xy + yz + zx = 1

sym
x
1 − x
2
=
4xyz
(1 − x
2
)(1 − y
2
)(1 − z
2
)
f(a, b, c) = |
|b − a|
|ab|
+
b + a
ab

−
2
c
| +
|b − a|
|ab|
+
b + a
ab
+
2
c
f(a, b, c) = 4max{
1
a
,
1
b
,
1
c
}
a
b + c
+
b
c + a
+
c
a + b

= 1
a
2
b + c
+
b
2
c + a
+
c
2
a + b
= 0
x + y
z + t
+
y + z
t + x
+
z + t
x + y
+
t + x
y + z
x
y + z + t
=
y
z + t + x
=

z
t + x + y
=
t
x + y + z
x + y = z + t
x
2
+ y
2
+ z
2
+ t
2
= (x + y)
2
+ (x − z)
2
+ (x − t)
2
ab + bc + ca = 1
(1 + a
2
)(1 + b
2
)(1 + c
2
) = [(a + b)(b + c)(c + a)]
2
a, b, c b = c a + b = c c

2
+ 2(ab − bc − ca) = 0
a
2
+ (a − c)
2
b
2
+ (b − c)
2
=
a − c
b − c
a
b − c
+
b
c − a
+
c
a − b
= 0
a
(b − c)
2
+
b
(c − a)
2
+

c
(a − b)
2
= 0
x, y, z
xy + yz + zx = 0, a =

y
2
+ yz + z
2
b =
√
z
2
+ zx + x
2
, c =

x
2
+ xy + y
2
(a + b − c)(b + c − a)(c + a − b) = 0
a, b, c, d ac + bd = (b + d + a −c)(b +
d − a + c)
(ab + cd)(ad + bc) = (ac + bd)(a
2
− ac + c
2

)
a, b, c a
2
+ b
2
+ c
2
= a
3
+ b
3
+ c
3
= 1
a + b
2
+ c
3
= 1
a, b, c, d a + b
2
= c + d
2
, a
2
+ b = c
2
+ d
a + b + c + d ≤ 2 {a, b} = {x, y}
a, b, c, d a + b + c + d =

a
7
+ b
7
+ c
7
+ d
7
= 0
(a + b)(a + c)(a + d) = 0

cyclic
a
k
(a − x)(a − y)
(a − b)(a − c)
=
xy
abc
•

cyclic
x(y + z)
2
− 4xyz = (x + y)(y + z)(z + x)
• 1 + x + x
2
+ x
3
+ x

4
+ x
5
= (1 + x)(1 + x + x
2
)(1 − x + x
2
)
• (ab + bc + ca)(a + b + c) − abc = (a + b)(b + c)(c + a)
• (1 + x + x
2
+ x
3
+ x
4
+ x
5
)
2
−x
5
= (1 + x + x
2
+ x
3
+ x
4
) ·(1 + x +
x
2

+ x
3
+ x
4
+ x
5
+ x
6
)
a+b(1+a)+c(1+a)(1+b)+···+l(1+a)(1+b) ···(1+k) = (1+a)(1+b) ···(1+l)−1
a = b = c = ··· = l
a + b + c = 0
(
b − c
a
+
c − a
b
+
a − b
c
)(
a
b − c
+
b
c − a
+
c
a − b

) = 9
a
2
k+1
− b
2
k+1
a − b
= (a + b)(a
2
+ b
2
)(a
4
+ b
4
) ···(a
2
k
+ b
2
k
)
a, b, c, d





a + 4b + 9c + 16d = 1

4a + 9b + 16c + 25d = 12
9a + 16b + 25c + 36d = 123
16a + 25b + 36c + 49d
a, b, c
a
b
=
b
c
=
c
a
a + b + c
a + b − c
a, b, c, d
a
4
b
+
b
4
d
=
1
b + d
a
2
+ c
2
= 1

a
2004
b
1002
+
b
2004
d
1002
=
2
(b + d)
1002
xyz = 1
1
1 + x + xy
+
1
1 + y + yz
+
1
1 + z + zx
= 1
2 +
√
3
√
2 +

2 +

√
3
+
2 −
√
3
√
2 −

2 −
√
3
=
√
2
3

3
√
2 − 1 =
3

1
9
−
3

2
9
+

3

4
9
A
a
=
B
b
=
C
c
=
D
d
√
Aa +
√
Bb +
√
Cc +
√
Dd =

(a + b + c + d)(A + B + C + D)
ax
3
= by
3
= cz

3
1
x
+
1
y
+
1
z
= 1
3

ax
2
+ by
2
+ cz
2
=
3
√
a +
3
√
b +
3
√
c
a, b, c, d abcd = 1
a + b + c + d =

1
a
+
1
b
+
1
c
+
1
d
1

4 −

10 − 2
√
5 −

4 +

10 − 2
√
5 = 1 −
√
5

2 +
√
3 +


14 − 5
√
3 = 3
√
2
3

6 +

847
27
+
3

6 −

847
27
= 3

4 −
√
15 +

5 +
√
21 +

6 −

√
35 +
√
6
3

7 +
8
3

55
3
+
3

7 −
8
3

55
3

5 +

17 + 2
√
7+

5 +


17 − 2
√
7+

5 −

17 + 2
√
7−

5 +

17 − 2
√
7
4

2 +
√
5 + 2

2 +
√
5 +
4

2 +
√
5 − 2


2 +
√
5
a, b, c
a
2002
=
b
2003
=
c
2004
4(a − b)(b − c) = (c − a)
2
x, y x
2
+ xy + y
2
= 0

x
x + y

2001
+

y
x + y

2001

n 2n + 1 {2, 5, 9}
a
1
, a
2
, . . . , a
2n+1
a
2n+1
= a
1
a
1
a
2
− a
2
a
3
+ ··· + a
2n−1
a
2n
− a
2n
a
2n+1
= 0
a, b, c ∈ R
1

bc − a
2
+
1
ca − b
2
+
1
ab − c
2
= 0
a
(bc − a
2
)
2
+
b
(ca − b
2
)
2
+
c
(ab − c
2
)
2
= 0
n n x

1
, . . . , x
n
k
S
k
= x
k
1
+ ··· + x
k
n
S
2
= S
3
= S
4
S
k
= S
1
k
x, y
(
√
x
2
+ 3 + x)(


y
2
+ 3 + y) = 1
x + y = 0
x, y, z xyz(x + y + z) = 1
(x + y)(y + z)(z + x) =
1
x
+
1
z
+ (x + z)xz
a, b, c, d, e, f





|d + e − a − b| =
√
3 ·

|b − a| + |e − d|

|e + f − b − c| =
√
3 ·

|c − b| + |e − f|


|f + a − c − d| =
√
3 ·

|c − d| + |f − a|

a + c + e = b + d + f
n w = cos
2kπ
n + 1
+ i · sin
2kπ
n + 1
n + 1 1 1
a
k
=

cos
2kπ
n + 1

n
1 + a
1
w + a
2
w
2
+ ··· + a

n
w
n
= 0
• cos a + cos b = 2 cos(
a+b
2
) · cos(
a−b
2
)
• cos a − cos b = −2 sin(
a+b
2
) · sin(
a−b
2
)
• sin a + sin b = 2 sin(
a+b
2
) · cos(
a−b
2
)
• sin a − sin b = 2 cos(
a+b
2
) · sin(
a−b

2
)
• tan a + tan b =
sin(a+b)
cos a·cos b
• cos a · cos b =
1
2
[cos(a + b) + cos(a − b)]
• sin a · cos b =
1
2
[sin(a + b) + sin(a − b)]
• cos(a + b) · cos(a − b) = cos
2
a − sin
2
b
• (cos a + cos b)
2
+ (sin a + sin b)
2
= 4 cos
2
a−b
2
• (cos a − cos b)
2
+ (sin a − sin b)
2

= 4 sin
2
a−b
2
• cos(a + b) = cos a · cos b −sin a · sin b
• cos(a − b) = cos a · cos b + sin a · sin b
• sin(a + b) = sin a · cos b + cos a · sin b
• sin(a − b) = sin a · cos b −cos a · sin b
• sin 2a = 2 sin a · cos a
• cos 2a = cos
2
a − sin
2
a = 2 cos
2
a − 1 = 1 − 2 sin
2
a
• cos
2
a + sin
2
a = 1
• tan 2a =
2 tan a
1−tan
2
a
• sin 3a = 3 sin a − 4 sin
3

a
• cos 3a = 4 cos
3
a − 3 cos a
• tan 3a =
3 tan a−tan
3
a
1−3tg
2
a
• tan a − tan b =
sin(a−b)
cos a·cos b
• cot a + cot b =
sin(a+b)
sin a·sin b
• cot a − cot b =
sin(b−a)
sin a·sin b
tan
a
2
= 4 tan
b
2
tan
a − b
2
=

3 sin b
5 − 3 cos b
a cos x + b cos y = a cos(x + z) + b cos(y + z) = 0 z = kπ
t ∈ R
a cos(x + t) + b cos(y + t) = 0
• tan
4
a =
cos 4a−4 cos 2a+3
cos 4a+4 cos 2a+3
•
1
2
· cot
4
a =
sin
2
2a+4 sin
2
a−4
1−8 sin
2
a−cos 4a
• cot a − tan a − 2 tan 2a − 4 tan 4a = 8 cot 8a
• cos
6
a − sin
6
a =

(3+cos
2
2a) cos 2a
4
• 2(sin
6
a + cos
6
a) − 3(sin
4
a + cos
4
a) + 1 = 0
•
1+sin 2a
sin a+cos a
−
1−tan
2
a
2
1+tan
2
a
2
•
√
1+cos a+
√
1−cos a

√
1+cos a−
√
1−cos a
= cot(
a
2
+
π
4
)
sin(a + 2b) = 2 sin a tan(a + b) = 3 tan b
sin(x − α)
sin(x − β)
=
a
b
cos(x − α)
cos(x − β)
=
a
1
b
1
ab
1
+ a
1
b = 0
cos(α − β) =

aa
1
+ bb
1
ab
1
+ a
1
b
f(x) = a sin x + b cos x x
1
, x
2
x
1
− x
2
= k · π k ∈ Z f(x
1
) = f (x
2
) = 0
f(x)
a = cos(x −α), b = sin(x−β) a
2
−2ab sin(α−
β) + b
2
= cos
2

(α − β)
x = 2y x+y+z = π (sin y+sin z)·sin y = sin
2
x
0 < α, β <
π
2

3 sin
2
α + 2 sin
2
β = 1
3 sin 2α −2 sin 2β = 0
α + 2β =
π
2
r
2
− 1
1 + 2r cos u + r
2
=
1 + 2r cos v + r
2
r
2
− 1
r
2

− 1
1 + 2r cos u + r
2
=
r + cos u
r −cos v
= ±
sin u
sin v
= −
1 + r cos u
1 + r cos v
tan
u
2
· tan
v
2
= ±
r + 1
r −1
cos x = tan y, cos y = tan z, cos z = tan x
sin x = sin y = sin z =
√
5 − 1
2
sin x
a
1
=

sin 3x
a
3
=
sin 5x
a
5
a
1
+ a
5
a
3
=
a
3
− a
1
a
1
cos x
a
1
=
cos 2x
a
2
=
cos 3x
a

3
sin
2
x
2
=
2a
2
− a
1
− a
3
4a
2
sin β
sin(2α + β)
=
m
n
1 +
tan α
tan β
m + n
=
1 − tan α tan β
m − n
0 < α, β <
π
2
α = β

cos x − cos α
cos x − cos β
=
sin
2
α cos β
sin
2
β cos α
tan
2
x
2
= tan
2
α
2
· tan
2
β
2
sin(a + b − c − d) =
sin(a − c) sin(a − d)
sin(a − b)
+
sin(b − c) sin(b − d)
sin(b − a)

cyclic
tan a −

sin(a + b + c)

cyclic
cos a
=

cyclic
tan a

cyclic
sin a − sin(a + b + c) = 4

cyclic
sin
a + b
2

cyclic
cos a − cos(a + b + c) = 4

cyclic
cos
a + b
2
sin
4
α
a
+
cos

4
β
b
=
1
a + b
sin
8
α
a
3
+
cos
8
β
b
3
=
1
(a + b)
3
a, b, c

cyclic
sin(a − b)
cos a · cos b
= 0

cyclic
sin a

sin(a − b) sin(a − c)
= 0

cyclic
cos a
sin(a − b) sin(a − c)
= 0

cyclic
sin a sin(b −c) cos(b + c − a) = 0

cyclic
cos a sin(b −c) sin(b + c − a) = 0

cyclic
1
sin(a − b)(a − c)
=
1
2

cyclic
cos
a−b
2
a + b + c = π

cyclic
sin 3a sin
3

(b − c) = 0

cyclic
sin 3a cos
3
(b − c) = 0

cyclic
sin
3
a cos(b − c) = 0

cyclic
sin
3
a sin(b − c) = 0

cyclic
sin a = 4

cyclic
cos
a
2

cyclic
cos a = 1 + 4

cyclic
sin

a
2

cyclic
tan a =

cyclic
tan a

cyclic
tan
a
2
tan
b
2
= 1

cyclic
sin 2a = 4

cyclic
sin a

cyclic
cos
2
a = 2

cyclic

cos a + 1
a
1
cos α
1
+ a
2
cos α
2
+ ··· + a
n
cos α
n
= 0 a
1
cos(α
1
+ θ) +
a
2
cos(α
2
+ θ) + ··· + a
n
cos(α
n
+ θ) = 0 θ = k · π(k ∈ Z)
λ ∈ R
a
1

cos(α
1
+ λ) + a
2
cos(α
2
+ λ) + ··· + a
n
cos(α
n
+ λ) = 0
a
tan(α + x)
=
b
tan(α + y)
=
c
tan(α + z)

cyclic
a + b
a − b
sin
2
(x − y) = 0
(a+b) sin(x−α) = (a−b) sin(x+α ) atg
x
2
= c+b tan

α
2
sin α =
2bc
a
2
− b
2
− c
2
tan 3α = tan α ·tan(60
0
− α) · tan(60
0
+ α)
cot 3α = cot α ·cot(60
0
− α) · cot(60
0
+ α)
tan
2
α + tan
2
(60
0
− α) + tan
2
(60
0

+ α) = 9 tan
2
3α + 6
sin 18
0
=
√
5 − 1
4
tg15
0
= 2 −
√
3
sin 15
0
=
√
6 −
√
2
4
cos 15
0
=
√
6 −
√
2
4

cos 18
0
=
1
4

10 + 2
√
2
sin 6
0
=

30 − 6
√
5 −

6 + 2
√
5
8
cos 6
0
=

18 + 6
√
5 +

10 − 2

√
5
8

cos 42
0
+ cos 102
0
+ cos 114
0
+ cos 174
0

2
=
3
4
tan 3
0
·tan 17
0
·tan 23
0
·tan 37
0
·tan 43
0
·tan 57
0
·tan 63

0
·tan 77
0
·tan 83
0
= tan 27
0
cos
π
15
· cos
2π
15
· cos
3π
15
· cos
4π
15
· cos
5π
15
· cos
6π
15
· cos
7π
15
=
1

128
−1 < x < 1
6

k=0
1 − x
2
1 − 2x cos
2kπ
7
+ x
2
= 7 ·
1 + x
7
1 − x
7
1
sin
2
π
7
+
1
sin
2
2π
7
+
1

sin
2
3π
7
= 8
tan
2
5
0
+ tan
2
10
0
+ ··· + tan
2
80
0
+ tan
2
85
0
= 195
tan 7
0
30

=
√
6 +
√

2 −
√
3 − 2
tan
2
1
0
+ tan
2
2
0
+ ··· + tan
2
89
0
= 4005
tan
6
20
0
− 33 tan
4
20
0
+ 27 tan 20
0
= 3
cos
4
π

14
+ cos
4
3π
14
+ cos
4
5π
14
= 12 cos
2
π
14
· cos
2
3π
14
· cos
2
5π
14
cos
8π
35
+ cos
12π
35
+ cos
18π
35

=
1
2
cos
π
5
+
√
7
2
sin
π
5
a
1
, a
2
, . . . , a
n
+1 −1
2 sin

a
1
+
a
1
a
2
2

+
a
1
a
2
a
3
4
+···+
a
1
a
2
···a
n
2
n−1

45
0
= a
1

2 + a
2

2 + ··· + a
n
√
2

tan(a + b) = 3 tan a sin(2a + 2b) + sin 2a = 2 sin 2b
1
cos 290
0
+
1
√
3 sin 250
0
=
4
√
3
tan 9
0
− tan 63
0
+ tan 81
0
− tan 27
0
= 4
cos 10
0
cos 50
0
cos 70
0
=
√

3
8
tan
π
11
+ 4 tan
2π
11
=
√
11
tan
6
20
0
+ tan
6
40
0
+ tan
6
80
0
= 33273
tan
6
10
0
+ tan
6

50
0
+ tan
6
70
0
= 433
4 cos 36
0
+ cot 7
0
30

=
√
1 +
√
2 +
√
3 +
√
4 +
√
5 +
√
6
cos
π
12
· cos

9π
12
+ cos
9π
12
· cos
17π
12
+ cos
17π
12
· cos
π
12
= −
3
4
1995

k=1
tan n tan(n + 1) =
tan 1996
tan 1
− 1996
n

k=1
cos
2
2k ·π

2n + 1
=
n
2
−
1
4
 = cos
2π
n
+ i sin
2π
n
n
A
k
= a
0
+ a
1

k
+ a
2

2k
+ ··· + a
n−1

(n−1)k

k = 0, 1, 2, , n − 1
n−1

k=0
|A
k
|
2
= n{a
2
0
+ a
2
1
+ ··· + a
2
n−1
}
cos nφ
cos
n
φ
= 1 −

n
2

tan
2
φ +


n
4

tan
4
φ − ··· + A
A = (−1)
n
2
tan
n
φ n (−1)
n−1
2

n
n−1

tan
n−1
φ
sin nφ
cos
n
φ
=

n
1


tan φ −

n
3

tan
3
φ +

n
5

tg
5
φ − ··· + A
A = (−1)
n−2
2

n
n−1

tg
n−1
φ n (−1)
n−1
2
tan
n

φ n
0 < α ≤ π 0 < β ≤ π
cos α + cos β −cos(α + β) =
3
2
α = β =
π
3
0 < α ≤ π 0 < β ≤ π
cos α · cos β ·cos(α + β) = −
1
8
α = β =
π
3
cos θ +cos ϕ = a; sin θ + sin ϕ = b
cos(θ + ϕ) =
a
2
− b
2
a
2
+ b
2
sin(θ + ϕ) =
2ab
a
2
+ b

2
α β a cos x + b sin x = c
cos
2
α − β
2
=
c
2
a
2
+ b
2
cos α = cos β = cos γ ·cos θ; sin α = 2 sin
ϕ
2
sin
θ
2
tan
2
α
2
= tan
2
β
2
tan
2
γ

2
(x − a) cos θ + y sin θ = (x − a) cos θ
1
+ y sin θ
1
= a
tan
θ
2
− tan
θ
1
2
= 2l
y
2
= 2ax − (1 − l
2
)x
2
x cos θ + t sin θ = x cos ϕ + y sin ϕ = 2a
2 sin
θ
2
· sin
ϕ
2
= 1
y
2

= 4a(a − x)
cos θ = cos α · β
tg
θ + α
2
· tg
θ −α
2
= tan
2
β
2
cos x
a
=
cos(x + θ)
b
=
cos(x + 2θ)
c
=
cos(x + 3θ)
d
a + c
b
=
b + d
c
α, β sin
2

α + sin
2
β =
sin(α + β) α + β =
π
2
cos
2
θ =
cos α
cos β
; cos
2
ϕ =
cos γ
cos β
;
tan θ
tan ϕ
=
tan α
tan γ
tan
2
α
2
· tan
2
γ
2

= tan
2
β
2
cos θ = cos α cos β cos ϕ = cos α
1
cos β tan
θ
2
·
tan
ϕ
2
= tan
β
2
sin
2
β = (
1
cos α
− 1) · (
1
cos α
1
− 1)
x cos(α + β) + cos(α − β) = x cos(β + γ) + cos(β − γ) =
x cos(γ + α) + cos(γ − α)
tan α
tan

1
2
(β + γ)
=
tan β
tan
1
2
(α + γ)
=
tan γ
tan
1
2
(α + β)
sin(θ −β) · cos α
sin(ϕ − β) · cos β
+
cos(α + θ) · sin β
cos(ϕ − β) · sin α
= 0
tan θ ·tan α
tan ϕ · tan β
+
cos(α − β)
cos(α + β)
= 0
tan θ =
1
2

(tan β + cot α); tan ϕ =
1
2
(tan β − cot α)
n
2
sin
2
(α + β) = sin
2
α + sin
2
β − 2 sin α sin β cos(α − β)
tan α =
1 ± n
1 ∓ n
tan β
cos(θ −α) = a; sin(θ −α) = b
a
2
− 2ab sin(α − β) + b
2
= cos
2
(α − β)
cos(α − 3θ) = m cos
3
θ sin(α − 3θ) =
m sin
3

θ
m
2
+ m cos α = 2
cos θ =
sin β
sin α
, cos ϕ =
sin γ
sin α
, cos(θ −ϕ) = sin β · sin γ
tan
2
α = tan
2
β + tan
2
γ
tan α, tan β
x
2
+ πx +
√
2 = 0
sin
2
(α + β) + π sin(α + β) cos(α + β) +
√
2 cos
2

(α + β) =
√
2
a sin
2
θ + b cos
2
θ = a cos
2
ϕ + b sin
2
ϕ = 1 ; a tan θ = b tan ϕ
(a − b)(a + b − 2ab) = 0
x
2n
− 1 = (x
2
− 1)
n−1

k=1
(x
2
− 2x cos
kπ
n
+ 1)
x
2n+1
− 1 = (x − 1)

n

k=1
(x
2
− 2x cos
2kπ
2n + 1
+ 1)
x
2n+1
+ 1 = (x + 1)
n

k=1
(x
2
− 2x cos
2kπ
2n + 1
+ 1)
x
2n+1
+ 1 =
n−1

k=0
(x
2
− 2x cos

(2k + 1)π
2n
+ 1)
sin
π
2n
· sin
2π
2n
···sin
(n − 1)π
2n
=
√
n
2
n−1
cos
2π
2n + 1
· cos
4π
2n + 1
···cos
2nπ
2n + 1
=
(−1)
n
2

2
n
sin
π
2n + 1
· sin
2π
2n + 1
···sin
nπ
2n + 1
=
√
n
2
n
cos
π
2n + 1
· cos
2π
2n + 1
···cos
nπ
2n + 1
=
1
2
n
cos

π
2n
· cos
2π
2n
···cos
(n − 1)π
2n
=
√
n
2
n−1
tan
π
2n
· tan
2π
2n
···tan
(n − 1)π
2n
= 1
tan
π
2n + 1
· tan
2π
2n + 1
···tan

nπ
2n + 1
=
√
2n + 1
cos α+i sin α x
n
+p
1
x
n−1
+
···+ p
n
= 0 p
1
sin α + p
2
sin 2α + ···+ p
n
sin nα = 0 p
1
, p
2
, . . . , p
n
3

cos
2π

7
+
3

cos
4π
7
+
3

cos
8π
7
=
3

1
2
(5 −
3
√
7)
3

cos
2π
9
+
3


cos
4π
9
+
3

cos
8π
9
=
3

1
2
(3
3
√
9 − 6)